Fourth power of a prime.

The fourth power of any prime greater
than 5 is one more than a multiple of 120.

I stumbled across that by noting (p-2)(p-1)(p+1)(p+2)
must be divisible by 120 since one of the brackets must
be a multiple of 5, one must be a multiple of 3, and of the two
even numbers one must be a multiple of 4.
Hence the product is divisible by 5*3*2*4 (=120)

By expanding the brackets and playing around one
can then show that p^4 -1 is a multiple of 120.

But there is a much easier proof.
What is it?

One form of Fermat's Little Theorem is:
p^(n-1) - 1 = nm
(p, m, n all integers, p not divisible by n)

So in your case we have
p^4 - 1 = 5m

Now p^4 - 1 is the difference of two squares:
p^4 - 1 = (p^2 - 1)(p^2 + 1)
And p^2 - 1 is also the difference of two squares:
p^2 - 1 = (p - 1)(p + 1)

Since p is a prime larger than 5, p must be odd and so both p - 1 and p + 1 are divisible by 2 and one of them is divisible by 4 (consecutive even numbers).

Also, by considering the three consecutive numbers (p-1), p, (p+1), it is clear that either p-1 or p+1 must be divisible by 3 (since p can't be).

Hence (p-1)(p+1) is divisible by 2x4x3 = 24.

Hence p^2 - 1 is divisible by 24.

Hence (p^2 - 1)(p^2 + 1) is divisible by 24

Hence p^4 - 1 is divisible by 24

We already know that p^4 -1 is divisible by 5 (the Fermat's Little Theorem bit at the start), and since 24 and 5 are co-prime, we know that p^4 - 1 is divisible by 5x24 = 120, i.e. p^4 = 120m + 1 for some integer m.

@mynameisklint
Well done!
(I did it without help from Fermat ... 😉 )